Boolean Algebra: What are the steps for simplifying this?

[HamsterEater]

(I hope this is the correct forum...)
Help me understand how I go about simplifying this:

RC'G'M'+RC'G'M+RC'GM'+RC'GM+RCG'M'+RCG'M+RCGM'

R, C, G, and M are just any given Boolean values, in this case.

 

[tkhunny]

Re: Boolean Algebra: What are the steps for simplifying this

You've almost a complete collection, there.

1) R is just R in every term. Factor it out.
2) After factoring, the only thing missing from the set is CGM, the ultimate AND gate.
3) Since you've everything except AND, it must be NOT AND.

Did I convince you?

 

[HamsterEater]

Re: Boolean Algebra: What are the steps for simplifying this

I think I see. So then it's just R(C'G'M')?


This doesn't seem like a solution that was algebraically found, though. I mean, I can memorize a new rule that all but one of the combinations of the variables must mean it is NOT the remaining one, but is that really the only way to discover this?

 

[tkhunny]

Re: Boolean Algebra: What are the steps for simplifying this

No. It's R(CGM)'.

It's a little tough to define what's algebraic and what isn't. Just my opinion. If your instructor has something particular in mind, you should do that, first.

RC'G'M'+RC'G'M+RC'GM'+RC'GM+RCG'M'+RCG'M+RCGM'

= R(C'G'M'+C'G'M+C'GM'+C'GM+CG'M'+CG'M+CGM')

= R(C'(G'M'+G'M+GM'+GM)+C(G'M'+G'M+GM'))

= R(C'(G'(M'+M)+G(M'+M))+C(G'(M'+M)+G(M')))

= R(C'(G'(M'+M)+G(M'+M))+C(G'(M'+M)+G(M'+M)-GM))

M' + M = 1

= R(C'(G'+G)+C(G'+G-GM))

= R(C'(G'+G)+C(G'+G)-CGM)

G' + G = 1

= R(C'+C-CGM)

C' + C = 1

= R(1-CGM) = R(CGM)'

How was that for algebraic?